Q21.111CP - Chemistry: Molecular Nature Of Matter And Change

Question

Car manufacturers are developing engines that use ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}$ as fuel. In Iceland, Sweden, and other parts of Scandinavia, where hydroelectric plants produce inexpensive electric power, the ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}$ can be made industrially by the electrolysis of water.

(a) How many coulombs are needed to produce $\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{·}{\mathbf{10}}^{\mathbf{6}}\mathbf{ }\mathit{L}$ of ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}$ gas at  $\mathbf{\text{12.0 atm}}$ and  ${\mathbf{\text{25}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$? (Assume the ideal gas law applies.) 

(b) If the coulombs are supplied at  $\mathbf{\text{1.44 V}}$, how many joules are produced? 

(c) If the combustion of oil yields $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{ }\mathbf{KJ}\mathbf{/}\mathbf{Kg}$ , what mass of oil must be burned to yield the number of joules in part (b)?

Step-by-Step Solution

Answer

(a) The amount of charge needed to produce $3.5\times {10}^{6}L$ of $H_2$ gas is $3.31·{10}^{11}C$ .

(b) The amount of energy produced, if the coulombs are supplied at $1.44V$ is $4.766·{10}^{11}J$.

(c) The mass of oil must be burned to yield $4.766·{10}^{11} J$energy is $1.19·{10}^4 Kg$ .

Step 1: Concept Introduction

The process of dissolving ionic compounds into their constituent components by delivering a direct electric current through the complex in a fluid form is known as electrolysis. At the cathode, cations are reduced, whereas anions are oxidised.

The equation of state of a hypothetical ideal gas is known as the ideal gas law. Although it has significant drawbacks, it is a good approximation of the behaviour of various gases under many conditions.

Step 2: Information Provided

• Hydrogen is a product of reduction of water, so, the reduction half-reaction is –

          ${\text{2H}}_{\text{2}}{\text{O + 2e}}^{\text{ - }}\to {\text{H}}_{\text{2}}{\text{ + 2OH}}^{\text{ - }}$

• Find the charge needed to produce of $3.5·{10}^6 L$ gas at 12.0 atm and ${\text{25}}^{\text{o}}{\text{C (25}}^{\text{o}}\text{C = 25 + 273 K = 298 K)}$
• Faraday constant: charge of 1 mole of electrons ($\text{F = 96485 C/mole}$)

Step 3: Calculation for Charge

(a) First, find the number of moles of $H_2$ produced –

$\begin{array}{rcl}\mathrm{PV}& =& \mathrm{nRT}\\ \mathrm{n}& =& \frac{\mathrm{PV}}{\mathrm{RT}}\\ & =& \frac{12.0 \mathrm{atm}·3.5·{10}^6}{0.082057 \mathrm{L} \mathrm{atm} {\mathrm{mol}}^{-1^{}}{\mathrm{K}}^{-1}·298\mathrm{K}}\\ & =& 1.7176·{10}^{6 }\mathrm{mol} {\mathrm{H}}_2\end{array}$

Since 2 moles of electrons produce of $H_2$, the number of moles of electrons is –

$\begin{array}{rcl}\mathrm{Moles} {\mathrm{e}}^{-}& =& 1.7176·{10}^6 \mathrm{mol} {\mathrm{H}}_2·\frac{2 \mathrm{mol} {\mathrm{e}}^{-}}{1 \mathrm{mol} {\mathrm{H}}_2}\\ & =& 3.45·{10}^{6 }\mathrm{mol} {\mathrm{e}}^{-}\end{array}$

Calculate the charge using Faraday constant –

$\begin{array}{rcl}\mathrm{Charge}& =& {\mathrm{mol}}^{-}·\mathrm{F}\\ & =& 3.435·{10}^6 \mathrm{mol} {\mathrm{e}}^{-}·96485\frac{\mathrm{C}}{\mathrm{mol} {\mathrm{e}}^{-}}\\ & =& 3.31·{10}^{11} \mathrm{C}\end{array}$

 

Therefore, the value for charge is obtained as $3.31·{10}^{11}C$.

Step 4: Calculation for Energy

(b) The coulombs are supplied at 1.44V, then find the energy produced –

$\begin{array}{rcl}\mathrm{potential}& =& \frac{\mathrm{energy}}{\mathrm{charge}}\\ \mathrm{energy}& =& \mathrm{potential} \mathrm{charge}\\ & =& 1.44\mathrm{V}·3.31·{10}^{11}\mathrm{C}\\ & =& 4.766·{10}^{11}\mathrm{J}\end{array}$

 

Therefore, the value for energy is obtained as $4.766·{10}^{11}J$ .

Step 5: Calculation for Mass

(c) The combustion of oil yields $4.0·{10}^4 \mathrm{kJ}/\mathrm{kg}=4.0·{10}^7\mathrm{J}/\mathrm{kg}$

The mass of oil that needs to be burned is –

$\begin{array}{rcl}\mathrm{m}& =& \frac{4.766·{10}^{11}\mathrm{J}}{4.0·{10}^7 \mathrm{J}/\mathrm{kg}}\\ & =& 1.19·{10}^4 \mathrm{kg}\end{array}$

 

Therefore, the value for mass is obtained as $1.19·{10}^4 \mathrm{kg}$ .